How to define a specific ring using a homomorphism If we have a ring $R$ then I can form a ring of matrices isomorphic to $R$ by setting $r \overset{\phi}{\mapsto}  
\left( \begin{array}{ccc}
r & 0 \\
0 & 0 \end{array} \right) $ and defining my ring of matrices to be the image of $\phi$. However, if I want to define this rigorously, I must specify which sets $\phi$ maps between, so I say $\phi:R \to M_2(R)$. However now $\phi$ is not technically a ring homomorphism as it does not map the identity of $R, \, (1)$ to the identity of $M_2(R)\,\, (=I_2)$ (though it does map to $1$ to the identity of the ring I want to form). So is it just impossible to define this ring as the image of a homomorphism without defining the ring already? I think I can define the set $$\tilde R=\left( \begin{array}{ccc}
R & 0 \\
0 & 0 \end{array} \right) = 
\left\{ \left( \begin{array}{ccc}
r & 0 \\
0 & 0 \end{array} \right):r \in R \right\}$$ 
and then give this a ring structure by defining $\phi : R \to \tilde R$, but is there no better way?
Edit I'm finding it difficult to phrase exactly what my question is, so I apologise for being unclear. My question is how I construct the specific matrix ring I give above, not just any matrix ring isomorphic to $R$. Though I'm aware that the construction @RobertLewis gives obviously works, this is a different (but isomorphic) ring. The reason I ask the question is that I want some 'neat' way of defining my ring above, but doing it via homomorphisms seems tricky. It's weird that $\phi:R \to M_2(R)$ isn't a ring homomorphism, but $\phi:R \to \tilde R$ is, even though $\tilde R \subset M_2(R)$ as sets. My question is basically to get to the heart of this, as it seems deeply unintuitive/wrong to me, even though it looks formally correct.
 A: The only alternative I can think of would be to consider everything as a non-unital ring, a.k.a. "rng" (Wikipedia link). Then the map $\phi:R\to\mathrm{M}_2(R)$ is a perfectly fine rng homomorphism, and since  the rng $R$ happens to have an identity, so will the image $\phi(R)\subset\mathrm{M}_2(R)$.
If you want to stick with rings instead of rngs, I would have to say "yes, there is no better way than to just define $\tilde{R}$".
A: I'm a little unclear about the phrase "define this ring as the image of a homomorphism without defining the ring already"; but nevertheless:
What if we map $r \to rI$; that is, define a map
$\theta:R \to M_n(R); \;\; \theta(r) = rI, \tag{1}$
where $I \in M_n(R)$ is the $n \times n$ identity matrix?  Then
$\theta(r + s) = (r + s)I = rI + sI = \theta(r) + \theta(s) \tag{2}$
and
$\theta(rs) = (rs)I = (rI)(sI) = \theta(r) \theta(s); \tag{3}$
$\theta$ is clearly a ring homomorphism.  Also, 
$\theta(1) = 1I = I; \tag{4}$
$\theta$ maps the identity of $R$ to the identity of $M_n(R)$, as requested.  Also, if
$\theta(r) = 0, \tag{5}$
then
$rI = 0, \tag{6}$
which clearly implies $r = 0$; $\theta$ is in fact in injective homomorphism into a subring of $M_n(R)$; $R$ is, via $\theta$, isomorphic to the subring $RI$ of $M_n(R)$.
You can define a subring of $M_n(R)$ to be the image $R$ under $\theta$.
How's that?
